proof of Ingham Inequality
Let
f(x)=√12πn∑j=-ncjeitjx. |
If k(x) is a integrable function in (-∞,∞), let
K(u)=∫∞-∞k(x)eixu𝑑x. |
It is easy to prove the equalities.
∫∞-∞k(x)|f(x)|2𝑑x | = | n∑j,h=-ncj¯chK(tj-th). | ||
∫∞-∞k(x)f(x)e-ixth𝑑x | = | n∑j=-ncjK(tj-th). |
For the rest of the proof we make the following choices:
k(x)={cos(x/2),|x|≤π0,|x|>π |
Then, after computation, we have
K(u)=4cos(πu)1-4u2. |
Let T=π and γ=1+δ (δ=ε/π>0). Since |tj+1-tj|≥γ, we have that |tj-th|≥|j-h|γ. If we suppose that h is fixed, we have
∑j≠h|K(tj-th)| | = | ∑j≠h|4cosπ(tj-th)1-4(tj-th)2|≤∑j≠h44(j-h)2γ2-1≤ | ||
≤ | 4γ2∑j≠h14(j-h)2-1<8γ2∞∑r=114r2-1= | |||
= | 82γ2∞∑r=1(12r-1-12r+1)=4γ2=K(0)γ2. |
But, since 2|cj¯ch|≤|cj|2+|ch|2 and K(tj-th)=K(th-tj), we find
∫∞-∞k(x)|f(x)|2𝑑x | = | n∑j,h=-ncj¯chK(tj-th)= | ||
= | ∑j|cj|2K(0)+∑j,h;j≠h𝒪(j,h)|cj|2+|ch|22|K(tj-th)| | |||
= | ∑j|cj|2(K(0)+∑h≠j𝒪(j,h)|K(tj-th)|), |
where |𝒪(j,h)|≤1. Using the definition of k and the previous inequality, we have
∫π-π|f(x)|2𝑑x≥∑j|cj|2K(0)(1-1γ2), |
and we have obtained the conclusion.
Title | proof of Ingham Inequality |
---|---|
Canonical name | ProofOfInghamInequality |
Date of creation | 2013-03-22 15:55:18 |
Last modified on | 2013-03-22 15:55:18 |
Owner | ncrom (8997) |
Last modified by | ncrom (8997) |
Numerical id | 8 |
Author | ncrom (8997) |
Entry type | Proof |
Classification | msc 42B05 |